Important Formulas Coordinate Geometry - Quantitative Aptitude for SSC CGL

What is Coordinate Geometry?

Coordinate geometry is the branch of geometry in which points on a plane are described by ordered pairs of numbers called coordinates. Each point is written as (x, y), where x denotes the horizontal distance from the y-axis and y denotes the vertical distance from the x-axis. Using coordinates, geometric problems are converted into algebraic equations and solved using algebraic methods.

Coordinate Geometry: Basic Concepts

• The origin is the point where the x-axis and y-axis meet. Its coordinates are (0, 0).
• Positive x-values lie to the right of the origin and negative x-values lie to the left.
• Positive y-values lie above the origin and negative y-values lie below.
• The plane is divided into four quadrants numbered I, II, III and IV in the usual anti-clockwise order. Points in each quadrant have characteristic signs for (x, y): I(+,+), II(-,+), III(-,-), IV(+,-).

Formulas Required for Solving Coordinate Geometry Questions

• Distance between two points A(x₁, y₁) and B(x₂, y₂).

• Slope (gradient) of a line through two points (x₁, y₁) and (x₂, y₂) is given by the ratio of change in y to change in x.

• Slope from a linear equation ax + by = c. The slope m = -a/b.

• Midpoint of the line segment joining A(x₁, y₁) and B(x₂, y₂) is the average of coordinates.

• Section (internal division) - the coordinates of a point R(x, y) that divides AB internally in the ratio m : n (A to R : R to B) are

• Section (external division) - the coordinates of a point R(x, y) that divides AB externally in the ratio m : n are

• Centroid of triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is the average of the vertices' coordinates.

• Area of a triangle with vertices A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) can be found by the determinant (shoelace) formula:

\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

• Division of a line segment by a point If a point P(x, y) divides the join of A(x₁, y₁) and B(x₂, y₂) in the ratio m : n, then use the section formula (internal or external) as given above.

• Equation of a line in slope-intercept form is y = mx + c, where m is the slope and c the y-intercept. The point-slope form for a line with slope m through (x₁, y₁) is:

\[ y - y_1 = m(x - x_1) \]

Some Examples Using Above Formulas

Q1: Given two points A(2, 5) and B(6, 9), find the Distance between points A(2, 5) and B(6, 9) Ans:

Compute the differences in coordinates.

\[ \Delta x = x_2 - x_1 = 6 - 2 = 4 \]

\[ \Delta y = y_2 - y_1 = 9 - 5 = 4 \]

Apply the distance formula.

\[ \text{Distance} = \sqrt{(\Delta x)^2 + (\Delta y)^2} \]

\[ \text{Distance} = \sqrt{4^2 + 4^2} \]

\[ \text{Distance} = \sqrt{16 + 16} \]

\[ \text{Distance} = \sqrt{32} = 4\sqrt{2} \]

Q2: If the slope of a line is -2/3 and it passes through the point (2, 5), find the equation of the line in point-slope form. Ans:

Use the point-slope form with m = -2/3 and point (x₁, y₁) = (2, 5).

\[ y - y_1 = m(x - x_1) \]

\[ y - 5 = -\tfrac{2}{3}(x - 2) \]

Simplify to slope-intercept form if required.

\[ y - 5 = -\tfrac{2}{3}x + \tfrac{4}{3} \]

\[ y = -\tfrac{2}{3}x + \tfrac{4}{3} + 5 \]

\[ y = -\tfrac{2}{3}x + \tfrac{4}{3} + \tfrac{15}{3} \]

\[ y = -\tfrac{2}{3}x + \tfrac{19}{3} \]

Q3: A rectangle has vertices at points A(1, 1), B(5, 1), C(5, 3), and D(1, 3). Determine its area and perimeter. Ans:

Compute the side lengths using distance formula.

\[ \text{Length} = \text{distance between }B(5,1)\text{ and }C(5,3) \]

\[ \text{Length} = \sqrt{(5-5)^2 + (3-1)^2} = \sqrt{0 + 4} = 2 \]

\[ \text{Width} = \text{distance between }A(1,1)\text{ and }B(5,1) \]

\[ \text{Width} = \sqrt{(5-1)^2 + (1-1)^2} = \sqrt{16 + 0} = 4 \]

Area = Length × Width.

\[ \text{Area} = 2 \times 4 = 8 \text{ square units} \]

Perimeter = 2(Length + Width).

\[ \text{Perimeter} = 2(2 + 4) = 12 \text{ units} \]

Q4: Determine the equation of the circle with center C(2, -1) and a radius of 5. Ans:

The standard equation of a circle with centre (h, k) and radius r is

\[ (x - h)^2 + (y - k)^2 = r^2 \]

Substitute (h, k) = (2, -1) and r = 5.

\[ (x - 2)^2 + (y - (-1))^2 = 5^2 \]

\[ (x - 2)^2 + (y + 1)^2 = 25 \]

Q5: Given the points A(3, 6), B(8, 6), and C(5, 2), find the area of the triangle ABC. Ans:

Use the shoelace (determinant) formula for the area of a triangle:

\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Substitute (x₁, y₁) = (3, 6), (x₂, y₂) = (8, 6), (x₃, y₃) = (5, 2).

\[ \text{Area} = \frac{1}{2} \left| 3(6 - 2) + 8(2 - 6) + 5(6 - 6) \right| \]

\[ \text{Area} = \frac{1}{2} \left| 3 \times 4 + 8 \times (-4) + 5 \times 0 \right| \]

\[ \text{Area} = \frac{1}{2} \left| 12 - 32 + 0 \right| \]

\[ \text{Area} = \frac{1}{2} \times 20 = 10 \]

Note: The algebraic steps above give the absolute value of -20; correct evaluation yields

\[ \text{Area} = \frac{1}{2} \left| -20 \right| = 10 \text{ square units} \]

Summary (optional)

The key formulas-distance between two points, slope, midpoint, section formulas (internal and external), centroid, area of triangle, equation of line and circle-allow translation of geometric problems into algebraic computations. Apply the appropriate formula, substitute coordinates carefully, and follow algebraic simplification step by step to reach correct answers.